如何在表单中使用序列化方法 [英] How do I use serialize method in my form
问题描述
我如何在我的表格中使用序列化方法
SERIALIZE（）;
我尝试过：
这是我的html表格： -
< html>
< head>
< title> VIGNESH DESIGNS.INC< / title>
< script src =http://ajax.googleapis.com/ajax/libs/jquery/1.11 .0 / jquery.min.js>< / script>
< script src =script.js>< / script>
< ; / head>
< body>
< form>
VIGNESH DESIGNS.INC
填写你的信息并获得放置
< label>名称：< / label>
< input id =nametype =text>
< label>电子邮件：< / label>
< input id =emailtype =text>
< label>密码：< / label>
< input id =passwordtype =password>
< label>联系号码：< / label>
< input id =contacttype =text>
&l t;输入id =submittype =buttonvalue =提交>
< / form>
< / body>
< / html>
这是我的PHP代码： - $ connection = mysql_connect（localhost，root，）;
$ db = mysql_select_db（vicky，$ connection）;
$ name2 = $ _ POST ['name1'];
$ email2 = $ _ POST ['email1'];
$ password2 = $ _POST ['password1'];
$ contact2 = $ _ POST ['contact1'];
$ query = mysql_query（插入vignesh_inc（姓名，电子邮件，密码，联系）值（'$ name2'，'$ email2'，'$ password2'，'$ contact2'））;
echo表格已成功提交;
mysql_close（$连接）;
？>
这是我的jqery ajax代码： -
$（document）.ready（function（）{
$（＃submit）。click（function（）{
var name = $（＃ name）。val（）;
var email = $（＃email）。val（）;
var password = $（＃password）。 val（）;
var contact = $（＃contact）。val（）;
//当输入的信息存储在数据库中时，返回成功的数据提交消息。
var dataString ='name1 ='+ name +'& email1 ='+ email +'& password1 ='+ password +'& contact1 ='+ contact;
if（name ==''|| email ==''|| password ==''|| contact ==''）
{
警告（请填写所有字段）;
}
其他
{
// AJAX代码到提交表格。
$ .ajax（{
类型：POST，
url：ajaxsubmit.php，
数据：dataString，
cache：false，
成功：函数（结果）{
al ert（结果）;
}
}）;
}
返回false;
}）;
}）;
how do i use serialize method in my form
SERIALIZE () ;
What I have tried:
this is my html form :-
<html>
<head>
<title>VIGNESH DESIGNS.INC</title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<script src="script.js"></script>
</head>
<body>
<form>
VIGNESH DESIGNS.INC
Fill Your Information And get placed
<label>Name :</label>
<input id="name" type="text">
<label>Email :</label>
<input id="email" type="text">
<label>Password :</label>
<input id="password" type="password">
<label>Contact No :</label>
<input id="contact" type="text">
<input id="submit" type="button" value="Submit">
</form>
</body>
</html>
this is my php code :- $connection = mysql_connect("localhost", "root", "");
$db = mysql_select_db("vicky", $connection);
$name2=$_POST['name1'];
$email2=$_POST['email1'];
$password2=$_POST['password1'];
$contact2=$_POST['contact1'];
$query = mysql_query("insert into vignesh_inc(name, email, password, contact) values ('$name2', '$email2', '$password2','$contact2')");
echo "Form Submitted Succesfully";
mysql_close($connection);
?>
and this is my jqery ajax code :-
$(document).ready(function(){
$("#submit").click(function(){
var name = $("#name").val();
var email = $("#email").val();
var password = $("#password").val();
var contact = $("#contact").val();
// Returns successful data submission message when the entered information is stored in database.
var dataString = 'name1='+ name + '&email1='+ email + '&password1='+ password + '&contact1='+ contact;
if(name==''||email==''||password==''||contact=='')
{
alert("Please Fill All Fields");
}
else
{
// AJAX Code To Submit Form.
$.ajax({
type: "POST",
url: "ajaxsubmit.php",
data: dataString,
cache: false,
success: function(result){
alert(result);
}
});
}
return false;
});
});
推荐答案
connection = mysql_connect（localhost，root，）;
connection = mysql_connect("localhost", "root", "");
db = mysql_select_db（vicky，
db = mysql_select_db("vicky",
connection）;
connection);
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